
Baby Analysis with Musings
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| Last Edited by | The Grand Inquisitor |
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Index for Homework and Quizzes
| Index | HW 1 | HW 2 | HW 3 | HW 4 | Qz 1 | Qz 2 | Qz 3 | Qz 4 | Eamples |
|---|---|---|---|---|---|---|---|---|---|
| 1 | See/ | See/ | See | See | 1,2 | See | See | ||
| 2 | See/ | See | See | See | See* | ||||
| 3 | See/ | See | 1/,2,3 | See | See | See | |||
| 4 | See | 1, 2,3 , 4 | See | See | See | ||||
| 5 | 2,3,4 | See | See | See | See | ||||
| 6 | See | See | |||||||
| 7 | See | See | |||||||
| 8 | 1, 2 | See | |||||||
| 9 | See | See | |||||||
| 10 | 1,2,3,4 | See | |||||||
| 11 | 1,2 | See | |||||||
| 12 | See |
0. 驯龙集
变脸法:缝朝服,至公卿 (take 的操作)
- Prove: UF-1 Standard definition of .
By UF-1, Now take , and standard defn holds.
- Prove of Fact 1.5
- Hw1 T3 Back to index

(i) false. Take .
(ii) take , take .
(iii) take . Let the largest index of the finitely many be .
let .
- Prove: UF-1 Standard definition of .
变脸法:公卿在哪 (take 的操作)
- Hw1 T4 奇偶子列分别收敛于同一值,证原序列也收敛到那个值 Back to index

Take
- Qz1 T3 奇偶子列都,证原序列也去天堂 Back to index
Take .
- Hw3 T3.2 易错:的取值只能依赖于, 不能依赖于, 因为需要满足. Back to index

Counterexample: .
- Hw1 T4 奇偶子列分别收敛于同一值,证原序列也收敛到那个值 Back to index
中流寻砥柱
- Prove: UF-2 UF-1.
We only need to consider .
- Prove: UF-2 UF-1.
Triangle Inequality
- Prove:
Let
- Hw1 T2: Prove: convergence of convergence of . Back to index
Key: . Maybe not exactly triangle inequality but…
- Prove:
Divide and Conquer
Common ways to divide:
- Fundamental: 断点(the birthday hoax),点之前利用finite # of #取max,点之后利用极限的定义和triangle inequality.
- Proof of 1.1, Toeplitz Theorem,
- Hw1 T5.2,
Basic 断点
- Hw1 T5.2: Back to index


Alternatively,
Take on both sides,
Therefore, , hence there exists N such that , hence .
- Hw1 T5.2: Back to index
Telescoping:
Hw3 T4: (1) is convergent is convergent. (2) In particular, if is convergent, then is convergent. is convergent is convergent. Back to index
(1) 抓紧
“”
Let is convergent exists as a finite number. is convergent.
“”
is convergent exists as a finite number converges.
(2)
converges converges.
Hw3 T3 (iii): satisfies integer , prove that is Cauchy. Back to index
WLOG, let Notice that
But is convergent, hence
, as desired.
Common ways to conquer:
- Convergence boundedness, always
- 之前,finite, always ; 之后,always
你们一起上吧,我能打十个 (放缩求和的时候,多让它几项也无妨)
Rearrangement 1: If converges absolutely, then any rearrangement also converges absolutely. Moreover,
Let be a rearrangement, with partial sums . Given , there exists an integer such that implies
Now choose such that the integer are all contained in the set . Hence if , the numbers will cancel in the difference . After the elimination, all the elements left in the partial sum have a index greater than . Let and be the partial sums after eliminating common elements, then
But , hence we have
Hence ,
Rearrangement 2 (Hw4 T4): Back to index
Let be a convergent series. Suppose is a rearrangement of satisfying that exist fixed integer such that , , where . Prove that converges and equals .
Notice that in this case:
The first terms in the original sequence must cancel out in as for any we can find such that .
After elimination, all the elements remaining in the partial sums have a index greater than and less or equal to . Denote and be the partial sums after elimination, then
Since is convergent, by divergence test, we must have .
Hence as long as is bigger enough, then is bigger enough (because is a fixed constant), then , .
Therefore, , , thus ,
Prove quotient rule:
Since convergent sequence is bounded, we can find a , such that .
Since , then
, thus we have
We then want to show that is smaller than a constant, which is equivalent to showing is bigger than a constant. How do we do this. Observe that if is large enough, we have now all we need to do is make a constant, simple, we just take
Example 5: . Show that
Observe
Similarly,
Therefore,
We already know that , the first term is nice, but the second term is awful, we want to have a common factor.
Notice that converges implies that is bounded, hence
Consequently,
Hw1 T5.3 (Toeplitz Theorem): Suppose integer . and . Prove that if , is a finite number, then . Back to index
Proof of Fact 1.1 的“虚张声势”版,思路非常基础:断点然后分而治之。
Since , then .
Therefore,
Exploit the fact that and , we have
We want the first term to be less than . Using the fact that ,
Take , then ,
Then
Special case given as Example 2 in W1L2: Suppose
Hw1 T5.4: Back to index

key: let . Then try to prove 1) , 2) .
(最后再乘一个, 而)
证明 ,
(fixed k, let n go to infinity. 指数比阶乘大)
Hw3 T12: Add your own flavor Back to index (它在这)
- Suppose , and converges with . Prove that .
Solution family 1
Notice that
is the remainder of
Hence (F3.2 尾巴趋于0, or F3.0殊始同归)
By squeezing principle, , hence .
Solution family 2 (Cind’s favorito!)
为了让 出现在 左边,我们需要构造一串和使得他们的角标能让项的个数与 有关。又由于 , 所以需要构造 单减所以往前跑,若单增,要放大,就往后跑。
Let , then Therefore converges to 0.
- Suppose , and converges with . Prove that .
Hw4 T5 plus (Cesaro-convergence): If is a complex sequence, define its arithmetic means by . Back to index
(i) Construct a sequence which does not converge, although .
Let . Hence is is even and if is even. Thus though has no limit.
(ii) If , prove that . (Usual convergence implies Cesaro-convergence)
Since , then
Hence
Moreover, given is convergent, is bounded, hence .
Therefore, if .
Essentially the same with Example 2 and Hw1 T5 Toeplitz Theorem.
Can it happen that for all and that , although .
Let if is not a perfect cube and if is a perfect cube (Construct a bad guy). Then if we have
(为了形式好看多让了几项)
The first sum on the right tends to zero by part (a) applied to the sequence . As for the last term, since we have , which tends to zero as tends to zero as . Since , it follows that tends to infinity as tends to infinity, and hence we have , even though (hence ).
(iii) Put . Show that . Assume that and that converges. Prove that converges.
If , then , hence , and . Then apply part (a) to the sequence
we have , hence
(iv)
- Fundamental: 断点(the birthday hoax),点之前利用finite # of #取max,点之后利用极限的定义和triangle inequality.
Powerful 一言不合就上下极限
通过一定存在的东西()寻找不确定存在的东西()
- F2.5 (Useful trick)
Fix an arbitrary in a set, then do limit/sup/inf
Hw2 T5: Suppose . Prove that if and only if for any .Back to index
For any , take , then ,
By fact 2.3.1, there exists , s.t. as long as .
which implies .
hence . By squeezing principle, .
Alternatively, start from .
For any , take , then there exists such that
Hence , .
Take limsup on both sides, we have
Here comes our ole friend again: since for all , we have
@The Grand Inquisitor I like this . Ole friend: proof of theorem “root test more general than ratio test.”
Hw3 T7: m 进制表示法 Back to index
- Suppose and . Prove that exists (as a finite number).
- Hint: Fix an arbitrary integer then , integer and s.t. . Try to show .
Solution
Fix , then , . Hence
(这一步分出来挺关键的)
Take limsup on both sides, (since we don’t know whether exists, in these cases we often take lim sup/inf.)
(having an arbitrary something on the numerator and a heaven-going something on the demoninator comes really handy.)
Since , is bounded by . Hence both and converges to zero as goes to infinity, which implies that the RHS actually has a limit. . Alternatively, since .
Therefore, we conclude that .
Hence , since m is arbitrary.
Thus (the form of is often what we desire when we use these notations in the first place, 夹一个 出来!), which implies and exists. is due to monotonically increases.
系数堆积法()
一般是利用/构造等比数列, 时用于证convergence, 时用于证divergence.
Hw3 T5: Fixed point Back to index
- Suppose is differentiable on , satisfying , where is a constant belonging to . Fix an arbitrary . Define sequence as: . Prove that exists as a finite number , with . [Hint: Mean Value Theorem].
Solution
Notice that
To prove exists, it is sufficient to prove converges since absolute convergence implies convergence.
Then we compute by induction:
By induction
Hence , a finite number.
By comparison test, converges.
Hence exists & finite.
Root Test: if then converges
Take , then by fact 2.3.1, there exists such that as long as . Hence . Since as a geometric series converges, by comparison test, converges.
Ratio Test: if , then converges
Take , by ole result, there exists such that as long as is larger than . Hence
By induction:
Since converges, by comparision test, converges, hence converges.
Prove that Root Test is more general than Ratio Test
-
Proof
Denote and
Take , by ole result, there exists such that as long as n is larger than . Therefore,
By induction, we have
Hence
Take limsup on both sides (RHS actually has limit) we have
Since , , we have
The Bad Guy(s) - Proof by constructing/finding a special subsequence (Tian’s Favorite Trick)
抓住之后的东西:它是自由的。在一般的情况下是标准定义里的, 如果任务有变,则看变之后之后是啥。然后I have a dream分析之后推出需要take it to be啥,然后大功告成。
Formulating the rigorous definition of from naive version
The naive version is derived from taking the opposite of the definition for
The refined definition using the language of subsequence:
We now show the construction process of
(Remark: the key is that these indexes must be increasing)
Take , then there exists such that .
Relabel as .
To guarantee , we take , then there exist such that . Relabel as and we know .
Continue this process, we can have a sequence of bad guys.
Prove Fact 1.6.
- Fact 6: Suppose satisfies: subsequence of , sub-subsequence , s.t. exists & , where is independent of the choice of subsequence. Then exists & .
Proof by contradiction
Suppose otherwise (), then by the rigorous definition of , there exists a bad and a corresponding bad subsequence such that
Then for this subsequence, we cannot find a sub-sub sequence whose limit is . And how do we prove we cannot find such a subsubseq? Proof by contradiction again:
Under the assumption of the problem, a subsubseq, contradicting .
Example 6 MIDTERM - Would you like to be amazed? Back to index
Let be a bounded sequence satisfying , where is a constant such that . Show that .
Orthodox: A jumble of facts, extremely high illustrative value!
is bounded
@The Grand Inquisitor 小天,这个最后一个是用的哪个fact啊,I buy it but I don’t feel extremely certain…
@Cindy Zeng @The Grand Inquisitor ok….
Let
WTS , and use Fact 2 to prove .
subseq of that converges to . Let this subseq be , then is also a subseq of .
, similarly,
Similarly, .
Observe
Taking on both sides we have
Taking on both sides we have
If
Similarly .
If
Similarly . @The Grand Inquisitor @Cindy Zeng 哦哦,忘了前面条件是 了😭
Therefore we have .
@The Grand Inquisitor呜呜小天,我觉得我把自己绕进去了。怎么从以下四个式子直接推M=m=0? 我知道如果分类讨论c<-1 还是c>1 怎么得出来,但是如果不分开,直接一起,我咋推不出来了呢?
?
@Cindy Zeng 我觉得合一起是不好推,因为涉及不等式两边同乘c是否要变号的问题,而它前面的式子又搞不出来,这应该是police后来推广的强化版,原版c>1的话就没问题。 @Cindy Zeng 等等的话变号之后两个不等式都是, 那不能得出它们相减是的啊
@The Grand Inquisitor 换一个路径可以搞出来。我吃完饭回来写给你如果你没想出来的话。
Police: Proof by contradiction
Suppose , then there exists a bad and a corresponding bad subsequence such that for big enough .
On the other hand, by , we have
, which implies
Secret Room: I have a dream
Since and we can find a such that , with an appropriate choice of , we hope to construct an exploding subsequence in the form of .
To set off an explosion, we can choose where satisfies . A customary choice is , then .
Take and take , then we have
We relabel and we take , then we have
Continue this process, we can construct a subsequence where
and
But , hence is unbounded, thus violates the condition that is bounded.
Prove Fact 2.3.2: there always exists a subsequence whose limit is the limsup.
思考怎么抽子列,能抽出一个limit是原序列limsup的子列?也就是n足够大的时候,subseq的元素. 右边肯定满足,那怎么构建一个左边也满足的子序列?利用sup定义第二条就行。这个子序列不需要一开始就很接近,只要不断接近,最后无限接近,就好了。
Since ,
we have .
In the first round, we choose and take
Since , by the definition of sup, there exists a
Because . In addition, we also have , so we
In the second round, we choose , and let .
Similarly, we there exists a .
In the third round, we choose , and let
Then by the same logic, there exists a .
Continue this process, we can construct a subsequence
Since and is monotone decreasing,
we have , which implies
hence .
Prove: is Cauchy is convergent
The difficulty lies in formulating an explicit characterization of , since Cauchiness is ostensibly defined on “relative” conditions. (Now, isn’t this fascinating? You say something “relative,” but you imply something “definite.” I think what makes this possible is not the eye-catching small relative distance part () but the quieter, part - it allows a compression of long horizontal distances into short vertical distances, it renders potential “anchors” (like ), from which the tail would not sway too far.)
Comb: Firstly, we prove that is bounded, then by Bolzano-Weierstrass theorem, we can have a convergent subsequence (our first grasp of ). Since if is convergent, its subsequences should have the same limit, we then need to prove that the limit of the subseq is indeed the limit of the original sequence.
In definition of Cauchyness, take . Then there exists such that for all ,
Take , then we have .
So after , the sequence is bounded by , before , there are only finite elements hence must be bounded by its maximum. is bounded.
Then by BW theorem, subsequence . We then want to show .
Since , then .
Notice by triangular inequality, we can decompose as
The second term will be tiny given , we now use the Cauchyness to tackle the first term. Replace by , then from , we have . Now the key is to guarantee that . Since , then as long as is large enough (denote this threshold by ). Consequently, as long as
Thus .
In reflection, this decomposition is quite intuitive. We want to characterize the relationship between and something definite, , knowing the relative relationship between ’s. Therefore we decompose into a term that resembles a relative relationship and a term that ties the wanderer to his Heimat. Any wanderer. Is there a subsequence of wanderers whose Heimat we know definitely exist? BW becomes a natural try.
Prove
Denote . Since , we have .
Then we want to show , which is equivalent to show that , i.e. is larger than every element in .
Formally, , by definition, there exists a subsequence whose limit is . i.e., .
Notice that . 也下一楼来和它斗。
Then (Rationale: is a subsequence of , since , then any of its subsequence converges to ).
Consequently, , hence as desired.
Qz3P1 Primitive construction of a desired sequence Back to index
Suppose
is not Cauchy. Prove that , and subsequences and of , s.t.
And
Proof:
is not Cauchy So for Let … Let Continue this and we get the desired result.
没有什么难的啦!这道题很好地展示了最难的可能就是把新seq中的后一个元素和前一个元素建立合理的联系。
Prove the case in Root test
- If , then diverges
Proof
Unlike case 1, we may not find a such that as long as . Nevertheless, by fact 2.3.2, there exists a subsequence such that . We than want to show that diverges.
W.T.S:
Take (to make ), then there exists such that
Hence
Since , , we have
Therefore, diverges, diverges?
Or use the divergence test, since there is a subsequence goes to infinity, , hence by divergence test, diverges (这里推不出来,错的,老师上课讲的似乎有点疏漏就是最终他只证明了 diverges :) @The Grand Inquisitor and diverges. (因为, 所以由comparison test, diverges 可以推出 diverges )行!
But , hence diverges. @The Grand Inquisitor 是带有绝对值的,我上面打错了我的笔记里没有绝对值,不知道是我抄错了还是老师写错了。我的笔记也没有绝对值,可能是老师写忘了。
Hw3T1 The Happy-Face Parade Back to index
反证假设-多米诺传导-原条件不成立, 不算严格的bad guy, 是序列内部的传导和收放,但从“传导”的角度来说还是有相似之处的。
思路:反证,假设小于1,递推得到 让 m 趋于正无穷,得到.
Suppose Prove that
Proof: (By contradiction)
Suppose
By induction we can prove
But for any given is a finite number in a sequence and cannot be greater than .
先证其存,再求其值
这个比较方便的一点是,有很多东西可以用来证明极限(作为一个finite #)存在,且可能比找出确切值要容易。而一旦证明存在之后,又有很多办法来寻找极限的确切值,其中包括直接设 然后解一元方程。
Common Ways to Prove Existence of Limit
- MCT
- Boundedness Bolzano-Weierstrass original seq also converges
Common Ways to Find Limit (Given Existence of Limit)
- 得知一定存在之后,直接设 然后解一元方程。很多东西都是等于 的,比如子序列的极限,recursive form 最好解了。
W2L1 Example 3: does this sequence have a limit? If it does, what is it? Back to index
Solution:
It obviously monotonically increases. Now we prove it is bounded above by 2 using mathematical induction:
-
- assume , then
Therefore . Now by MCT, exists as a finite number. Observe .
Does exist? (The definition of ; proved by M.C.T.)
Proof see here.
Hw2 T2: Construct a sequence to approximate the root of . Back to index
- Fix a positive number . Choose , and define by the recursion formula . Prove that decreases monotonically and that .
Hw2 T3: 双生暗影 Back to index
- Fix . Take and define .
(a) Prove that ;
做差有点麻烦,考虑做商(注意到)。 即 where .
Compare and ,
if then , hence
if then , hence
Then all we need to show is .
Notice that is an decreasing function as .
Also, we have , hence
iff and iff .
Consequently, , , , …. as desired.
(b) Prove that ;
Given , we have .
(c) Prove that
By MCT, both and are convergent. Denote their limits as and respectively. We then utilize the recursion formula twice (at and ) to obtain two equations about .
At , take limit on both sides, we have
Similarly, at , we have
The solution is . Since both the even terms and odd terms converge to , the sequence converges to (by Hw1T4).
我觉得TA给的存后如何得其值的方法也挺elegant,值得借鉴的:做商。
.
Cauchy Criterion - 离间计
Cauchyness 的一大好处就是可以序列内部相减相消。这对于 这种加和样貌的序列很方便。
Example 8: Does exist? ( is bounded by and ) Back to index
, we have
(triangular inequality)
(boundedness of )
(sum of geometric series)
We want to show as long as is bigger enough. We solve for .
Take log on both sides,
In formal proof, we need to reverse the direction, and take
1. Limit of Sequence
Definitions
不论出身,不论生平,只论终相。然无终何以论终?身之棺具体,心之椁无形,可小之又小,小小不息。
-
Equivalently,
- (UF-1) What’s new:
Proof of equivalence: From standard to UF-1, just take , the other way around, see here.
Warning: cannot appear in !!! HW1P5 wrong original solution.
- (UF-2) What’s new: only talk about tiny teeny . 明显有最大值时考虑使用。
Proof of equivalence: From UF-1 to UF-2, obvious cuz UF-2 only talks about a subset of UF-1’s cases. From UF-2 to UF-1, see here.
- Example 1: Prove . Back to index
WTS
- Example 1: Prove . Back to index
- (UF-1) What’s new:
-
naive:
polished: Back to index
We now show the construction process of use the term
Note that he key is that both and goes to (partly by definition of subsequence).
Take , .
Relabel as .
To guarantee , we take , then there exist such that . Relabel as and we know .
Repeat this process, we have bad subseq .
-
-
第一个是极限存不存在的问题,等于一个 finite number 或者等于,都算是极限存在。极限不存在使用 就不是well-defined,所以我觉得也不是一个很严谨的说法,因为极限不存在的话是不能用这个符号的(就这样说有点默认的极限存在了,只是不等于罢了)。在的例子里,说它极限不存在我们也没有严谨证明。第二个问题是极限存在后,这个极限是不是有限的。
这种定义是把 是否convergent和converge到哪里混合在一起的,cauchy的定义就只涉及convergent,但不说converge到哪里。所以我想到的极限不存在的说法是 1. 不是柯西的 (这样它是divergent)+ 2. 不趋于.
行吧,应该不是big issue. 估计这几个terminologies都没有很统一,符号的意思大概比语言的意思更本质。

Basic Facts
1.1 收敛必有界
断点法首秀。点之前利用finite # of #取max,点之后利用极限的定义和triangle inequality.
Let
As for
Remark:
1.2 Squeezing Theorem: Suppose
只要足够大……那什么才叫足够大? 找 .
Proof:
1.4 极限与序列元素相对大小一致: Suppose for
Proof: (By contradiction) Suppose , we try to create a circumstance under which for large enough
. Dream: So we take
Corollary:
If
However, is WRONG.
1.5 子列与原列极限一致 Suppose Then for any subsequence , we have
Proof see here.
1.6 原列与子子列极限一致 Suppose satisfies: subsequence of , sub-subsequence , s.t. exists & , where is independent of the choice of subsequence. Then exists & .
Proof see here.
Remark: by 1.6, the converse of 1.5 holds: if for any subsequence , we have then
1.6 is more general than the converse of 1.5, because it only requires each subsequence to contain one () subsubseq that converge to , whereas the converse of 1.5 says any subseq itself should converge to . Note that sub-subsequence , s.t. exists & , doesn’t necessarily give limit of subseq also exists and is (e.g., ). However, if we include the part into our consideration, is 1.6 really more general than the converse of 1.5? Can we find a seq where “ subsequence of , sub-subsequence , s.t. exists & , where is independent of the choice of subsequence” is satisfied yet “ subsequence , we have ” is not? @The Grand Inquisitor
@Cindy Zeng 我觉得没有,因为如果前面的条件满足,那么 exists & , 然后由fact 5就可以推出后面那个条件。 所以这两个条件可以互相imply,它们应该是等价的。
@The Grand Inquisitor 我也觉得没有更general…
(1.7) Monotone Convergence Theorem
Suppose is monotone & bounded, then
- Supremum and Infimum
- A real # is said to be the “least upper bound (supremum 上确界)” of if
- is an upper bound of , i.e.,
-
Intuitively, either or elements of 在 之下紧密排布,“没有最近,总有更近的。”
- A real # is said to be the “greates lower bound (infimum 下确界)” of if
- is an lower bound of , i.e.,
-
Remark: upper bound of we have ; lower bound of we have .
Example: Hw2 T1 Back to index

(iv) rational number: can be represented by where p, q are integers. 思路和证明极限等于某某差不多。
- A real # is said to be the “least upper bound (supremum 上确界)” of if
- The Least Upper Bound Axiom (公理)
If is bounded from above, then exists. 总存在最小的上界。
Theorem: If is bounded from below, then exists.
Proof: Let is bounded from above. By axiom, exists as a finite number, and it satisfies
-
-
By-product: 名字相对,负号两次
- Proof of MCT
Only consider the case of increasing . is bounded from above exists, we let it be . By definition:
-
- 会给挤压到天花板上。
Remark: If & not bounded from above, (you’re going to heaven); If & not bounded from below, (you’re going to hell). Meaning if we adopt the conceptual framework described in the beginning of this note, monotonicity limit must exist (either as a finite number or as infinities).
- Example
Example 4: Does exist? (The definition of ; proved by M.C.T.)
- Supremum and Infimum
2. Upper and Lower Limits
Motivation
does not always exist, but upper and lower limites always exist. And we can use them to prove the existence of / directly find . Prof. Wang loves it. Powerful stuff.
Definition
Consider , WLOG, assume for convenience (we don’t really care about ).
And in this way we construct a new sequence and . By MCT, exists and may be . We call the “lower limit” of , and we denote it as or .
Similarly, we can construct a new sequence and , where . We call the “upper limit” of , and we denote it as or .
Basic Facts
2.1 下极限小于上极限
Proof:
2.2 中原存则蛮夷附之 exists (allow )
Proof:
“”.
case 1: (finite #)
case 2:
case 3: . Similar to case 2.
“”.
case 1: (finite #)
case 2:
case 3: . Similar to case 2.
2.3.1 上下极限之外,远则不遇
2.3.2
总存在收敛到上极限/下极限的子序列 Define a set .
2.3.3
(上)下极限是E的(上)下确界 Proofs:
2.3.1
Rest similar.
2.3.2
See here.
2.3.3
By Fact 2.3.2 we know that By definition of
2.4 上下极限与序列元素相对大小一致(类比F1.4)
Proof: @The Grand Inquisitor 小天,帮我看看这个证明对不对😭😫
Case 1: is a finite number.
Suppose , by Fact 2.3.1, , take . Let we can construct a bad sequence of where contradicting 到红色为止就可以结束了,前面加一个, 就能得出的矛盾 (这个只是在某个值上面满足,似乎不够)
我当时想了想好像不行,我们需要一个完整的序列来contradict some N 那个部分,因为 不行,或许比他大的some N 还是可以work的。嗯你说的对!本来以为真就是F1.4的翻版,一试发现似乎没那么简单O.o#.
我觉得利用 (集合中每个元素都更小,inf也更小),让然后就可以直接用fact 1.4 说 , 也即 @The Grand Inquisitor 天,我突然想到一个问题,我们并不知道 存在啊?(@Cindy Zeng 根据MCT, 和 都存在,所以说liminf 和 limsup总是有的)
诶,有道理。别走 @The Grand Inquisitor 可是那我那个复杂的方法对不对啊,感觉每次取一个不同的很心虚。上次好像就错过一次,因为不能反复取.
这里没问题,极限不存在和不cauchy的例子不能反复取不同的是因为都是针对的同一个bad ,但像证明有一个子序列极限为limsup的时候,第n轮的,每轮都是递减的。
嗯对,有时可以,有时不行。可能就要注意分清什么时候要一到底,什么时候只需要找到一个它就好了(可能当是条件的一部分的时候,找到一个就行,但当我们需要说明 balabala 的时候就必须一到底)。主要是之前我错的那个地方我也记不太清了呜呜,好像是某道作业题,没事,应该找得到。
2.5 (Useful trick) 很有用!尤其是当我们不知道lim是否存在的时候。
2.6 合租房比整租房扁 (True as long as does not appear on RHS.)
Proof (Hw2T4.2, 4.3): Back to index
规整和上确界≤打乱和上确界≤分开上确界(让我想起高中作文素材曾摘了一个形容独立性的重要性的美句哈哈哈,!找到了!“廊柱分立才能撑起庙宇。橡树和松柏也不能在彼此的阴影中生长。”——纪伯伦《先知》顺便看到了一句无关的,分享分享:“有人曾问过王劲松,荀彧为什么会相信曹操可以帮他匡扶汉室?王劲松用剧中的一句词作答:万古长夜当中,哪怕有一盏微弱的光芒,都会让你身不由己追随,至死方休。” @The Grand Inquisitor )军师联盟😭😭😭其实这个(包括下面那个)的证明思路有点像minmax。什么minmax? 为什么我第一反应是信息竞赛里的什么东西???证明零和博弈纳什均衡的一个定理
我不知道这个😓,但是我第一眼看到你这个证明的时候就觉得这个思路很有前途😁。

Let
.
2.7 脱帽无妨 If exists as a finite #,
Proof (Hw2T4.4): Back to index
还是关于上下极限和极限三者分析的老思路:筷子都夹住了,中间必有汤圆。
Preserve the notations used in proving Fact 2.6. .
2.8 首尾颠倒,负号两次
Proof (Hw2T4.1): Back to index
Let , by the byproduct of the least upper bound axiom, .
Theorem: Bolzano-Weierstrass Theorem
If is bounded, then it contains a convergent subseq.
Proof:
is bounded is finite. a subseq converging to .
3. Series
Definition: Given a sequence of , we call
- an infinite series
- partial sums of the sequence
- constitutes a new sequence: if exists as a finite number, then we say converges and define . Otherwise, we say diverges.
Basic Facts
3.0 开端不影响结局,殊始同归 and have the same convergence/divergence nature. (although the value may be different).
3.1 最后增量需趋于零,天下归心: If converges, then . Consequently, if . then diverges. ( 是 converge 的必要不充分条件).
Proof: Observe . Take limit on both sides:
.
3.2 只要足够远,绝对片面和就足够小 converges (WLOG, ), .
Proof: converges exists & finite is Cauchy.
Remark 整条尾巴都要足够扁 remainder shrinks to 0: If converges, then as .
Proof: . Take limits on both sides of the equation
Because is just a finite number
Example: 尾巴扁不下来
Hw3 T10(b): Suppose , and diverges. Prove that and deduce that diverges. Back to index
Note that
The last inequality holds because hence is increasing.
To prove that diverges, we verify the opposite of its convergent conditions
Take , then , we take , so that
Since . For any fixed , we have , so we can find a big enough, such that . This means .
Hw3 T11(a): Back to index

Since converges, by remark of F3.2, as is not Cauchy, therefore it diverges.
For Cauchy stuff, you can tinker with since they are behind
3.3 Absolute Convergence 左边更难conv(所以更强),因为取和的话绝对值消除了正负相抵的可能性: converges converges.
Proof: By Fact 2,
(By Triangular Inequality)
Then by Fact 2 again, converges.
Compare: F1.2: convergence of convergence of but not the reverse. 因为不取和则绝对值更易converge,它可以把和两岸的人团结成一股力量(Der Mensch ist ein Seil über einem Abgrunde.)
3.4 Comparison Test: Suppose some , then
- 内收 If converges, then so does .
- 外推 If diverges, then so does .
Proof
(i) WLOG, assume .
. Moreover, since converges, exists as a finite number, then is bounded, denote the bound by . Therefore, .
In addition, , is monotone increasing and bounded below by 0. Consequently, is monotonically increasing and bounded, by MCT, exists as a finite number, and thus converges.
(ii) Since diverges and is monotonically increasing, by MCT, we have . Then and diverges.
- Remark: If , then either converges to a finite number or .
Examples: 常见锚点p-series/geometric series/telescope sum, i.e., and .
Hw3 T8 (a): Back to index
Multiplying and dividing by , we find that
Since diverges (), hence diverges
Alternatively, since the sum telescopes, the th partial sum , which obviously tends to infinity.
Hw3 T8 (b): Back to index
Using the same trick above, we find that .
Since converges (), by comparison test, converges
Hw3 T9: Let , prove that the convergence of implies the convergence of [Hint: Comparison test and ] Back to index
Note that
Since converges, converges, we have converges.
By comparison test, converges
HW3 T10 (c): Suppose , and diverges. Prove that and deduce that converges. Back to index
Denote . Hence
Therefore, converges. By comparison test, converges.
Hw3 T10 (d): Construct a sequence such that diverges but
Let if is not a perfect square and if is a perfect square (the bad guy). Back to index
, hence diverges
But
Both and converges, hence converges.
@Cindy Zeng 你觉得这个地方的rearrangement 可能会有问题嘛,就把能开平方数和非平方数分别求和。 虽然这里 都是正数,如果有convergence也有absolute convergence, 但有点预设了你要证明的东西的感觉。
@The Grand Inquisitor 我觉得没有问题哇,只要找到一个反例即可,如果这个反例确实满足原序列diverge 新序列converge 那不就可以了?不过为什么要提absolute convergence?
@Cindy Zeng 我担心的是如果原序列不是absolute convergent的话,rearrangement不一定等于原序列的和。We are too young to know too much
@The Grand Inquisitor I don’t think this is a rearrangement though. Its order is preserved, it’s just that you group them when you do addition. Is it? @The Grand Inquisitor
@The Grand Inquisitor 天,老师最后讲rearrangement 的那个proof的时候,最后蓝笔才是正确版本对吗?相当于之前的黑笔版本就不要了?
Hw3 T11 (b): Back to index

. Divide both sides by we have the desired inequality. Since (telescoping), and converges. By comparison test, … converges.
3.5 Limiting Comparison Test: Suppose some , if
- 辅车相依,唇亡齿寒 : and converge/diverge at the same time
- : convergence of implies convergence of and divergence of implies divergence of
- : convergence of implies convergence of and divergence of implies the divergence of
Proof of (1) 白矛纯束:
Take , then we have
Thus . Then by Comparison Test, if converges, then so does , and so does . Otherwise, if diverges, then so does , and so does
Proof of (2):
Take follows from 3.4 Comparison Test.
Examples
Hw3 T10 (a): Suppose , and diverges. Prove that diverges. Back to index
Tian:
Proof by contradiction: Suppose converges, then by divergence test, . Notice that , hence implies that .
Take the ratio between and : .
Therefore, is a constant, and should diverge simultaneously, contradiction.
Cind:
Proof by contradiction:
Suppose conv
Comparison test (fact 3.4) doesn’t work.
Use limiting comparison test (fact 3.5) So must converge with , which contradicts with the setup.
- 3.6 Suppose . Then converges iff is bounded
- @The Grand Inquisitor 天,为啥我突然觉得这里有点问题, 咋办。我的意思是如果这条fact成立那不就意味着 sn convergent sn bounded, which is not true.?
- 这里的前提是, 所以是单增的,单增的话convergence和boundedness就是可以互推的 @Cindy Zeng 哦对,我傻了 @The Grand Inquisitor
3.7 Integral Test 平行宇宙: Suppose , is decreasing, and there exists a continuous and decreasing function defined on s.t. . Then converge or diverge at the same time.
Proof of Case 1: converges.
Since , is increasing, to prove converges, we only need to show is bounded.
Since exists & finite, we have is bounded(Convergence implies boundedness). Therefore, is bounded.
Proof of Case 2: diverges
Similarly, to prove diverges, we only need to show is unbounded.
The LHS approaches infinity as n goes to infinity, hence is unbounded.
Example 9: p-series , converges if diverges if (or, ). Back to index
Let , which decreases on , and
By Integral Test, converges if diverges if .
3.8 Root Test: Let
If , then converges
Take , Then
.
Since as a geometric series converges, then by comparison test, converges.
If , then diverges
Since , then there exists a subsequence such that (F2.3.2). Consequently, take (so small that ), there exists such that
as long as . Hence .
But , hence diverges.
If , inconclusive
Counterexample: diverges, converges. Yet both of their .
3.9 Ratio Test
3.9.1 If , then converges.
We can find and so that ,
Take ,
Take ,
By induction, we have
Then (infinite geometric series, which is finite, times a finite number).
Thus converges (by MCT), which implies converges (by Fact 3.1).
3.9.2 If then diverges.
This means is increasing. By MCT, exists (may be )
But , then . By divergence test, diverges
3.9.3 If , then diverges.
Follows from 3.9.2, by F2.3.1.
Theorem: Root Test is more general than Ratio Test
-
Prove
Let
If , then , done.
If , then take , then there exists ,such that ,
Hence
Thus . Take liminf on both sides (RHS has limits)
Since this fixed is an arbitrary element from , we have
Prove
Let .
If then .
If , then take , then there exists , such that ,
Thus . Take limsup on both sides,
Since this fixed is an arbitrary element in , we have
Corollary, if exists, then also exists and equals
Proof: Squeezing Principle
Example 10: Try root test and ratio test on series: Back to index
Root test:
Want to know
Let
By root test the series converges absolutely.
Ratio test:
Want to know
Ratio test fails.
Root test prevails.
Power Series:
- Central Question: For what does the power series converge or diverge?
Theorem: Let , Let .
- For , converges absolutely.
- For , diverges.
- Remark: The theorem does not apply to points where
@The Grand Inquisitor 天,可以把你这里老师讲does not apply 的时候举的例子那部分笔记拍给我一下吗,我这里记的interval of convergence 是[-1,1), 但看过程应该是(-1,1)啊?我的搞错啦!确实应该是[-1,1)吧
是converge的,由alternating test
嗯嗯对!谢谢小天。我刚刚才看到alternating series test. 才反应过来。
Proof
Let , then .
Use the Root Test,
By Root Test, if , then converges absolutely, this requires
Similarly, if , then diverges, this requires
Example 11
Theorem: the partial sums of are bdd. is decreasing and approaches zero. Then converges.
Preparation: Summation by parts
- Let . The for , we have
Proof:
之后进行拆分尽力往desired样子靠就行了。
Proof
, we have
(Remark: no absolute value before because is positive)
Since is bounded, we can find a such that . Then
. (Remark: This is called the telescoping series, as all the barriers between the beginning and the ending get killed) . Then we have
Again, no absolute value in the second term because is decreasing.
Since , then , thus
Hence converges.
换汤不换药:
Hw4 T3: If converges, and if is monotonic and bounded, prove that converges. Back to index
Rudin
We shall show that the partial sums of this series form a Cauchy sequence, i.e., given there exists such that if .
By summation by parts, we have
Divide and conquer: Denote and be a bound of .
Since , we can find big enough such that
We only need to worry the last term. Given the sequence is monotonic,
where by Cauchyness. Hence
Alternating Series Test: Suppose is decreasing and approaches zero. Then converges. @The Grand Inquisitor 你这里写错了,是吧
Additions & Multiplication of Series
Addition
Suppose both converges, then
(1) also converges.
(2) also converges.
Multiplication
Cauchy product: Give and . Define Cauchy product of these two series by , where .
Theorem: If converges absolutely, converges, then Cauchy product converges and equals , where
The second term approaches to our desired result ; part of the first term vanishes to zero as approaches to zero, but terms like is not tiny.
Now we proceed with the formal language,
Since , then , thus
In the first terms, ; in the remaining terms,
Road 1:
Since converges, is bounded, hence , we have
Take , then since , there exists , such that . Hence we get
Road 2:
take on both sides.
The Magic Number e
Example 4: Does exist? (The definition of ; proved by M.C.T.) Back to index
We prove the existence of limits by M.C.T.
Firstly, we show is monotonically increasing.
Recall the binomial theorem: where
Secondly, we show that is bounded from above
Now by MCT, the limit exists as a finite number, which is .
Example 7: Prove . (proved by limsup and liminf) Back to index
We do not even know whether exists or not, hence we can only use limsup and liminf to manipulate the inequalities.
Ideally, we want to show
Firstly, we show that
Take liminf on both sides of the inequality
But since , we have
Hence we showed that
Secondly, we show that
If we can show that , since , then we are done. To accomplish this goal, we only need to show that
Fix an arbitrary , for any , we have
Essentially, we are dumping the elements after . Since is a fixed finite number, we can take limits on both sides of the inequality
Since this fixed is arbitrary, we actually have .
Note: Taylor expansion: so
Example 11: What is the interval of convergence for . This is the Taylor series of . Back to index
Solution:
First try root test: , ugly, try ratio test (will come to learn that
Try ratio test: let . By ratio test, converges absolutely, . So interval of convergence: .
Hw3 T6: Prove that is convergent. (The limit is called Euler’s constant, denoted by ). Back to index
- Hints
- Use , . 第一个不等号证bdd+之后放缩, 第二个证mono。Reminds me of proof of integral test, any strictly decreasing would satisfy I guess.

- Show is monotone and bounded.
- Use , . 第一个不等号证bdd+之后放缩, 第二个证mono。Reminds me of proof of integral test, any strictly decreasing would satisfy I guess.
Proof
Denote
Then
Since , we have .
To prove is bounded, Notice that , hence
We want to make the RHS bigger, use , thus
(Remark1: we do not enlarge the first term because is not defined or is )
(Remark2: This trick shares the same spirit as the integral test)
- Hints
4. Basic Point-Set Topology
Topology, no trick, all conceptual problems
Definitions
Metric space
Let be a metric space. All points and sets mentioned below are understood to be elements and subsets of .
Neighborhood
- Let , a neighborhood of is a ball with radius . Later on, we can use “ is within a small neighborhood of ’’ to replace “ is very close to .”
Limit point
- A point is said to be a limit point of if contains a point with .
Closed set
- Let be the set of all limit points of . Then is said to be closed if
Interior point
- A point is said to be an interior point of if nbhd .
Open set
Let be the set of all interior point of . is said to be open, if every point in is an interior point of , i.e. .
Bounded set
is bounded if there is a real number and a point such that for all .
Clousure
- The closure of is defined to be
Dense set
E is dense in if every point of is a limit point of , or a point of (or both), i.e. .
Finite, Countable, and Uncountable Sets
- A set is said to be finite if either or .
A set is said to be countable if is infinite and all its elements can be listed as a sequence .
- is countable.
- the interval is not countbale.
- A set is said to be at most countable if is either finite or countable.
We say is equivalent to (denoted as ) if function such that is 1-1 and onto.
Basic facts
- is countable .
- A subset of a countable set is at most countable.
- If each is countable, then is also countable.
If each is countable, . Then is countable.
set of all rational # is countbale.
Basic Facts
Basic facts 1: Any nbhd is open
Just need to prove , is an interior point of .
Use as radius to construct , w.t.s. .
:
Hence , and is open.
Basic facts 2: is a limit point of iff , contains infinitely many points of (2); iff such that for and .
- (2)(1): obvious
(1)(2)
Argue by contradiction. Suppose not, then , such that contains only finite many point of .
Define , then contains no points and
Notice that the min exists due to finiteness.
(3)(1)
Since as ,
, such that if , we have .
Hence and . What if ? Since and , we can let be the desired point.
(1)(3)
Construct such a sequence. Since is a limit point of , then contains point .
Take contains point .
Next time, take , within a smaller neighborhood, we can find .
By induction, take , and we can find in such that , .
The last thing is to prove the sequence we construct converges to .
By induction, we can show that , hence , so .
Weaker version, endorsed by TA: .
Related fact: s.t. is dense in .
Basic facts 3: Let be the complement of which is defined by . Then is open iff is closed. (Usually, it is easier to prove openness).
W.T.S. . Argue by contradiction. Suppose and . Since is open, is an interior point of , then . This contains no point in , contradicting to .
W.T.S. is an interior point of . Suppose not, then , such that contains points in . These points can not equal as , so we have is a limit point of . But given is closed, implies that , contradicting to .
Basic facts 4 Unions and Intersections of Open(Closed) Sets:
Let be a family of open sets of , then is open.
some . Since is open, then nbhd . Thus is an interior point of .
Let be a family of closed sets of , then is closed.
Use fact 3, . is open by fact 3, and is open by fact4(i), hence is closed.
In (1), if is finite, then is open.
Given is finite, denote . Then .
Since is open, then
Now let , then , hence , done. Again, the existence of min requires finiteness.
In (2), if is finite, then is closed.
Use the same trick in proving (2). By fact 3,
. By (3) and fact 3, is open and is closed.
Basic facts 5 Closure:
Closure of is closed.
Use fact 3, w.t.s. is open.
, we have . Hence there exists a bad , such that does not contain a point . Hence .
We want to show , but
So we need to show , or equivalently, .
Argue by contradiction, suppose , then and . Since is a limit point of , then , we have contains a point in . We can choose so small that (this is doable because neighborhood is open), then would contain a point in , contradiction.
is closed.
is obvious because of . As for Since is closed, , we havd . But we also have , hence .
If is closed in and , then , i.e. is the smallest closed set containing .
.
Hw 5 Q1:
Hw 5 Q2: Let be subsets of a metric space: (i) If , prove that , for ; (ii) If prove that .
- Key argument: If is a limit point of , then it must be the limit point of some . Suppose not, then for each , exists , such that does not contain any point in . Take (minimum exists in finite many numbers), then does not contain any point in . Similar to the prove of is open and is closed.
Hw 5 Q3 : Let denote the set of all interior points of . (a) Prove that is always open. (c) If and is open, prove that . (d) Prove that the complement of is the closure of the complement of . (e) Do and always have the same interiors? (f). Do and always have the same closure?
- (a) , consider , then , hence every point of is an interior point of ; (d) Two-side inclusion.
Hw 5 Q4: Let be a metric space and . The boundary of is defined to be , contains a point of and a point of . The boundary is denoted as . Prove: (a) is closed. (b) . (c) Is it always true that a boundary point must be in .
- (a)Usually, it is easy to prove open (exist is easy than any). Consider proving is open. , , sucht that or . Consider , then , thus .
- (b) Two-side inclusion
Hw 5 Q5: Let be bounded from above. Prove that
- Connect the definition of sup and the definition of limit points.
5. Compact Sets
Definition: A subset of a metric space is said to be compact if every open cover of contains a finite subcovers. More explicitly, the requirement is that if is an open cover of , then there are finite many indices such that .
- Trick to construct an open covering: .
Compact set is closed
- Prove is open. Key construction: , since , we can find a small sucht that . But is an open covering of , hence exists finite many s.t. . Again, minimum exists for finite many numbers, we can find a small such that , hence is an interior point of .
Compact set is bounded (Hw6 Q3)
- Key construction: let be the neighborhood of with radius . Then is an open covering of .
- Useful fact, finite number has maximum.
Heine-Borel: Let , then is compact is closed and bounded (Hw 6 Q4)
- Induction by dimension: use Bolzano-Weierstrass in each dimension.
- Application: Rectangle in is compact (Hw 6 Q5)
Sequential compact Compact (Hw 6 Q1&Q2)
- A subset of metric space is said to be sequentially compact if , convergent subsequence , s.t. , as .
Hw6 Q1: Compactness Sequentially Compactness
- Case 1: The set is a finite set, i.e. and occurs infinitely many times in , we can take , then .
Case 2: The set is infinite (i.e. it contains infinitly many distinctive points). Let .
Subcase1: has a limit point. Use the fact that there is a sequence with distinctive values converges to this point. We need to guarantee the index is increasing.
Denote the limit point be . Then for and . Since , choose the smallest s.t. .
If , choose the smallest s.t. .
If , we drop ,…, must exist (Because contains infinitely many elements and only deletes finitely many elements), we can then choose the smallest , s.t. .
We can thus get and .
Subcase2: has no limit point. We show that this cannot happen (Use the property of compactness)
- Possibility 1:. Since is not a limit point of , then nbhd such that .
- Possibility 2: . Since is not a limit point of , then does not contain any point in , i.e. .
Obviously, is an open covering of , given is compact, then there exists a finite set such that . Hence . But is finite, so is a finite set, a contradiction.
Hw6 Q2: Sequentially Compactness Compactness
- A partial proof, the open covering must has the form of (the set is countable)
Argue by contradiction. Suppose no finite subcovering of can cover . Then let . Then is increasing as increases, , so (i.e. ). Consider this sequence of , given is sequentially compact, subsequence such that some .
Since is increasing, then is decreasing. So . So fixed , let , then for all larger s. Since as the union of open sets is open, is closed. Because there is a sequence converges to , then is either a limit point of or a point in (or both). Given is closed, we must have .
Contradict with .
Closed subset of compact set is compact (Hw 6 Q7 & Quiz )
- Use sequential compact
- Useful fact: is closed convergent sequence , we have . (Hw 5 Q1)
Let be a family of compact set of . Suppose finite subset of , , then .
Argue by contradiction. Suppose . Do complement: . Fix , then . Observe that is open (because is compact, is closed), so is an open covering of . Since is compact, then , s.t. . Do complement again: . Thus contradicting the assumption.
Let be a sequence of non-empty compact set such that (decreasing), Then .
Application: Nested Interval Theorem (Hw 6 Q6)
6. Connected Sets
- Definition: Let metric space , we say and are separated if and . We say is connected if is not a union of two nonempty separated sets, i.e., nonempty , s.t., (1) ; (2) and are separated.
Basic facts
Basic facts 1: Equivalent definitions
- is connected
- open sets , s.t. (1) ; (2) are nonempty; (3) .
- closed sets , s.t. (1) ; (2) are nonempty; (3) .
Proof (a) (c)
Argue by contradiction. Suppose closed , s.t. where and are nonempty and .
We can let , then and are both non-empty and . We only need to show and .
Since and is closed, by ole result, (because is the smallest closet set containing ). Similarly, .
So
Hence and are separated, contradicting with is connected.
Proof (c) (a)
Argue by contradiction. Suppose nonempty ,s.t. and . Take .
Then
and .
, contradicting to the assumption.
- (a) (b) see Hw 7 Q2.
- Basic facts 2: Empty set is connected.
Basic facts 3: If is connected, then any which is both open and closed must be either or . (Quiz 6)
Suppose open & closed and . Then is also open & closed and . Notice that , so is not connected.
Basic facts 4: Any interval is connected.
Basic facts 5: A convex set is connected. (Hw7 Q3)
Hw 7 Q1: Suppose metric space is connected. (i) Prove that is also connected. (ii)Find a counter-example showing that is not connected.
- Key construction: , argue is not entirely contained in .
Hw 7 Q2: Prove the equivalent definition of connected set using open set: metric space is connected open sets and in such that . and are nonempty, and .
- Key construction: “”: ; “” .
“”
Suppose such that and both non-empty; . Let , then both non-empty and . We want to show .
For any :
(1) . Since , this .
(2) but . Suppose this , then we should have . But given is a limit point of . Since , this . Hence is not an interior point of , contradict with is open. Thus we must have .
Similarly, we have , contradict with is connected.
“”
Suppose non-empty such that and .
Let .
Since and , we have .
Then
are both non-empty, and ; , contradiction.
Hw 7 Q3: Prove that a convex set is connected.
- Key construction: Take , let , , use and to reach the contradiction that is not connected.
- Useful fact:
Quiz 6: Let be a connected metric space. Suppose is both open and closed in . Prove that is either or .
- Key observation: . And use the fact that open then closed,vice versa
7. Limit of Functions
Definition: Let . We say the limit of as is a point , if , , then whenever .
- may not in , and in general has nothing to do with .
- If , don’t talk about , because no body in the domain is approaching it.
- , 去心邻域
Cauchy’s criterion
- We say that a real-valued function defined near satisfies Cauchy’s conition as if , such that , whenever .
- We say that a real-valued function defined on satisfies Cauchy’s condition as if such that as long as , , .
Basic facts
. (化整为零)
Since , then , as long as . Since , , then whenever
Hence as long as , we have , thus .
Suppose not. Then bad bad , but .
Take and let , then , hence and . But which implies .
- Remark: bad , , s.t. bad and bad sequence , and
Addition, Multiplication and Quotient Rules for limits of real & complex-valued function
8. Continuity of Functions
Definition: Let . We say is continuous at if: (1) ; (2) , s.t. whenever , .
- may or may not . If , then is automatically continuous at , because is isolated and nobody in the domain is approching it except itself. If , then is continuous .
- If is continuous at every , then we say is a continuous function.
- Main Convenience: & . That is, the transformation perserves the convergence.
Basic Facts
(Composite function): If is continuous at , and is continuous at , then is continuous at . In particular, if and are continuous functions, then so is .
Since is continuous at , we have
if
Since is continuous at , for this , there exists , s.t.
if
Combine them together, we have , , s.t whenever ,
we have , and consequently,. Hence is continuous at .
(Pre-image)A mapping of a metric space into a metric space is continuous on . if and only if is open in for every open set in .
Suppose is continuous on and is an open set in . We have to show that every point of is an interior point of . So, suppose and . Since is open, there exists such that . Since is continuous at , there exists such that if . Thus .
For all , , let which is open. Then is open. Since , there exists . Hence as long as , we have
Corollary: A mapping of a metric space into a metric space is continuous if and only if is closed in for every closed set in .
Since for every
Continuity and Compactness
If is continuous and is compact, then is also compact.
- Motivation: is it true that and ? In general, No.
Proof 1: Sequential Compact
We just need to show is sequentially compact, that is , exists a convergent subsequence such that .
Consider , we can back out a corresponding sequence . That is, there exists such that . Since and is compact, there exists a convergent subsequence such that . Since is continous,
Proof 2: Finite Open Covering
Let be an open cover of . Since is a continuous function, we have each preimage is open. Consequently, is an open cover of . Given is compact, there are finite many indices, say such that
Since for every , this implies that
(Extreme Value Theorem) If is continuous and is compact, then , .
Since is continuous and is compact, then is compact. Since , we have is closed and bounded. Hence . In ole Hw, if and is bounded, then . Hence .
If is continuous & 1-1 & onto & is compact, then is also continuous.
, w.t.s is continuous at .
If , then is isolated, is automatically continuous at (we can always find a proper such that the neighbor hood of only contain itself, so that for all . )
If , then we need to show . Suppose not, then bad and bad sequence such that but . Since is 1-1 and onto, then such that . (or equivalently, ). Since is compact, subsequence such that . Since is continuous, . But . Hence . But , contradict with .
- Remark: If not compact, then this fact may fail.
(Uniformly Continuous): If is continuous & is compact, then is uniformly continuous on .
- Definition: Let , we say is uniformly continuous on if whenever .
Proof 1: Finite Open Covering
Fix . Since is continuous, for each point , we can find a , such that
if
Then is an open cover of . Since is finite, there is a finite set of points in such that
We choose (The min exists due to finiteness, this is one point where finiteness of the covering, inherent in the definition of compactness, is essential). Now let be points in such that .
Since , such that .
Then
Hence
Proof 2: Sequential Compact
Suppose not, then bad such that bad pair but .
Take , and let , then we have
but
Since is compact, then subseq and (do they have the same index?) such that and .
and fly together,
, so .
Since is continuous, we should have , But , they approach the same limit,
, contradcit with .
Hw 8 Q3: Replace by . Show that is uniformly continuous in the following two scenarios:
is bounded.
exists as a finite # (Divide into and )
Hw 8 Q4: Let be real-valued, uniformly continuous on finite interval . Prove that and exists as finite #. (Proving the existence of limits without knowing the exact value——Cauchy’s criterion; Treating as a sequence.)
Hw 8 Q5: Let and be continuous on with . Prove that , s.t. .
Continuity and Connectedness
Let be continuous and be connected. Then is connected
Proof1: Original definition of nonempty separated set
Proof2: Equivalent definition with open set
Suppose not, then open such that
-
-
- , .
Let . Since is continuous and are open, are open. Observe , hence
.
Hence .
This implies that
Then we show and . Since , then . Hence .
can be proved similarly.Finally, we need to show . Suppose not, then , such that contradiction.
(Intermediate Value Theorem): Let be continuous. Let , . If , then , , s.t. .
- Corollary: .
Continuity and Dense
Hw 7 Q5: If is a continuous mapping , prove that
- Use definition.
Hw 7 Q7: Let and be continuous mappings of a metric space into a metric space , and let be a dense subset of . Prove that is dense in . If for all , prove that for all .
- Useful fact: is dense in . Combine with get full-fledged.
- Key construction: .
Hw 8 Q1: Prove that the Dirichelt’s function (equals 1 if is rational and equals 0 otherwise) is not continuous at any . (Hint: is dense, the equivalent definition of limit points)
Continuity and Convexity (Hw 8 Q10)
9. Discontinuity of Functions
Discontinuity of Monotone Functions
- If exist as finite numbers and not equal, or but , then we say that has discontinuity of 1st kind at .
Fact: Suppose is increasing, then and exists as finite numbers, More precisely, . Furthermore, if , then
Proof:
Denote . Then , s.t.
Since is increasing, then on , .
Hene , and ()
Hence exists & . Similarly, exists & .
To prove , only need to show
Take and , then
Fix , take sup,
Since is arbitrary, we have
Moral of the story: monotone functions have only simple discontinuities. Their continuities can only come from ``jumps''
Theorem: If is increasing on , then the set of discontinuous point of is at most countable.
Let
Define a mapping . (i.e. output an interval).
Define
Then consider the composite , essentially, for every point , we associate a rational number such that . We want to show this mapping is 1-1.
If , WLOG, let , then , hence
Hence . Hence is at most countable.
10. Differentiation
Differentiation of Multivariable & Vector-valued functions
Motivation: Generalize from single-variable real-valued function
The difficulty is that we can not have vector divided by a vector.
Step 1: single-variable real-valued function
If is a real-valued function with domain and if , then is usually defined to be the real number
provided this limit exists. Thus
where the “remainder” is small, in the sense that it converges to 0 faster than
We call the higher order infinitisemal of , denoted as . It is any function satisfying .
Note that the difference can be expressed as the sum of the linear function that takes to , plus a small remainder. We can therefore regard the derivative of at , not as a real number, but as the linear operator on that takes to .
Step 2: single-variable vector-valued function
Let us next consider a function that maps into . In that case, was defined to be that vector (if there is one) for which:
We can again rewrite this in the form:
where . The main term of the RHS is again a linear function of .
Thus if is differentiable at then is the linear transformation of that satisfies
- Let . We say is differentiable at if there exists such that
as is called the total derivative of at . Notation .
Alternatively, .
Proving differentiability : Play with .
- Remark: represents a function satisfying . The term inside indicates the decaying speed, not the direction. We always take about the limit when , only the function we compared with is different.
Chain rule
Let be differentiable at . Let be differentiable at . Then is differentiable at . Moreover,
Proof
Since is differentiable at , we have
as .
Since is differentiable at , we have
as .
Take , since is continuous at , then as
Hence
Continuous partial differentiability Total differentiability
It suffices to show componentwise.
that is
Use a trick similar to telescoping, we build the bridge to link and by creating a series of . (minus some term, then plus back). Each time, we eliminate the increment on one dimension, and use the MVT of a single varible to better characterize the resulting difference.
……
Applying MVT on ,we have
Since all the partial derivatives are continuous, the coefficient in front of goes to 0 as . Hence we have
Since , they are all , then we have
HW9 Q6
HW10 Q1: Suppose that is a real-valued function defined on open set , and that the partial derivatives are bounded in . Prove that is continuous in .
Hw10 Q2: Let open set . Suppose satisifies exists and is continuous and exists. Prove that is differentiable at .
Mean Value Theorem (Killer’s Instinct)
Rolle’ s Theorem (Baby MVT): Suppose is real-valued continuous on , differentiable on , . Then such that .
MVT(Lagrange): Suppose is real-valued continuous on , differentiable on , then such that . (某处的瞬时速度等于整段的平均速度)
- Corollary: Suppose real-valued and differentiable on , then (i) If on , then is increasing on ; (ii) If on , then is decreasing on . (iii) If on , then on .
MVT(Cauchy): Let & be real-valued continuous on , differentiable on , then , such that .
MVT for multivariable vector-valued functions: Suppose is differentiable on & assume is convex, and is bounded on , denoted a bound by , then .
Proof
Trick: Transform a multivariable function into a single variable function
Fix arbitrary ,
( represents the dot product), this is a single variable real-valued function. Apply MVT to on
, for some
(Inequalities used: )
Take , then
- Corollary:If on and is convex, then we can take and get on .
Mixed derivatives
Let suppose exists & is continuous on . Then the mixed derivative also exists &
Lemma: Let be real-valued & defined near and (finite number). Assume , exists as a finite number, denoted as . Then .
Idea: 一维一维地逼近
.
Since , then
whenever
Now fix
Then for all close to . To obaint , sending
Since exists , such that as long as , we have .
WLOG, assume is a function of .
What we know:
Want to show: ..
Use MVT to create a similar form as what we know, eliminate
Method 1
. But may not equal .
Method 2
Alternatively, define
Then
Now the equation becomes:
Since is continuous, we have
.
L’ Hospital’s Rule (Hw9 Q2)
Suppose and are real-valued and differentiable on with and finite .
- If , then .
- If , then .
Properties of
(First Derivative Test)Suppose is an interior point of and is local extreme and is partially differentiable at , then .
(Hw9 Q3): The derivative function of a real-valued function also satisfies the Intermediate Value Property: Suppose is differentiable on with . Then between and , such that .
(Hw9 Q4): cannot have simple discontinuities in .
(Hw9 Q5): Let be differentiable and convex. Prove that is continuous on . (First show is increasing).
(Hw9 Q7): Let open and connected . Suppose .Prove that is constant in . (By Multivariable MVT, we have this result holds for convex sets)
(Hw9 Q11): If is differentiable on and if is increasing on , then is convex on .
(Hw10 Q3): Suppose , is a twice-differentiable real function on , and are the least upper bounds of , respectively, on . Prove that .
(Hw10 Q4): Suppose is twice-differentiable on is bounded on , and as . Prove that as .
Taylor’s Theorem
- Motivation: Define to be the order Taylor expansion at , which is a order polynomial that shares the first order derivatives with at . We want to use this polynomial to approximate (Not only locally around !!!). Then what can we say about the ?
Taylor’s Theorem with Cauchy’s remainder (Global result): Let . Suppose exists & continuous on and exists on , and . Then
for some between and . We can interpret as a higher order MVT.
Proof
Define , then we want to show .
Define . Notice that
Hence we need to show there exists a between and such that .
通过 次MVT, 在 到 之间搭桥。
Also since and share the same first order derivatives, we have
Hence . Our choice of gives .
Apply MVT to on , .
Apply MVT to on .
…….
Apply MVT on on , , as desired.
Taylor’s Theorem with Peano’s remainder (Local result, usually used in quotients): Suppose exists for some . Then
where represents any function such that .
Proof
W.T.S.
At , the numerator is zero, the denominator is also zero, 0/0, use L’Hospital
At , still 0/0, use L’Hospital again:
Repeat this process until we hit
Remark: we do not use L’Hospital to derive is because may not be continuos.
(Hw10 Q5):
(Hw10 Q6) Newton’s Method:
(Hw10 Q7) Taylor’s Theorem for functions of several variables:
(Hw10 Q8) Hessian matrix and second derivative test:
Implicit Function Theorem
Let be open in and be -smooth. Write a generic point in as . Suppose , and is non-singular. Then open sets , s.t.
-
-
Let is called the function implicitly defined by . satisfies:
-
- is -smooth
-
- .

[數學分析] Inverse Function TheoremChung-Han Hsieh's Bloghttps://ch-hsieh.blogspot.com/2014/12/inverse-function-theorem.html
[數學分析] 隱函數定理Chung-Han Hsieh's Bloghttps://ch-hsieh.blogspot.com/2012/11/implicit-function-theorem.html
隐函数定理22312;微积分中, 隐函数定理表明在适当的条件下 ...https://www.bananaspace.org/wiki/%E9%9A%90%E5%87%BD%E6%95%B0%E5%AE%9A%E7%90%86
11. Riemann Integral
Goal
- Given a set , how to rigorously define .
- Necessary and sufficient conditions on so that exists.
- Generalization of Fubini’s Theorem (iterated integral).
Starting point: is a rectangle.
Jargons
- Let be a closed rectangle in , then . and volume of .
- A partition of is a collection of cutting points . Intervals between two cutting points are the subintervals determined by .
- Let be a partition of . We say n-tuple is a partition of . is divided into tiny sub-rectangles in the following manner: If is a subinterval determiend by , then is called a sub-rectangle determined by , write as . The norm of equals the largest size of all sub-rectangles determined by .
(Riemann integrable): Let be real-valued and bounded on . We say is Riemann integrable on if as long as , we have , for some constant .
- We write .
- Then is called the integral of on , write as .
- Agony: don’t know the value of Using limsup and liminf to squeezes .
(Lower sum and upper sum): The choice of in each sub-rectangle is arbitrary. We can hence define the lower bound and upper bound of the Riemann sum as follows
-
-
- Let and be two partitions of , define the common refinement at . Then we have and .
(Lower integral and upper integral):
- Upper integral:
- Lower integral:
Fact1 (Hw12 Q1): is R-integrable on whenever .
Fact2: is R-integrable on partition such that . (does not require ).
Fact3: Let be bounded on . Then is R-integrable on is continuous almost everywhere on
- Intuition: Consider a sub-rectangle , if is continous on , then is almost the same for within as long as is small enough, hence difference between the sup and inf is negligible. For those containing discontinuous points, the difference between the sup and inf may be large(but still bounded), but their area is negligible.
Basic measure theory (define continuous almost everywhere).
- A set is said to have “measure” zero (denoted as ) if , sequence of closed rectangles in such that (i) (ii) .
Basic facts: (i) If and , then . (ii) If , then .
- Corollary (a countable set has measure zero) : If , then , and . In particular, the set of rational numbers has measure zero.

- We say is continuous almost every where on if the set of discountinous point has measure zero.
(Fubini’s Theorem on ): Let be closde rectangle in and cloased rectangle on . Let . Write generic point in as , where and .Suppose exists and , exists, then exists and equals .
General , use closed rectangle to cover
Properties of
@The Grand Inquisitor 哈哈小天,我在听最后两节课的录播,他说math research is much easier, you just need a pen and paper. 突然觉得做数学应该和做哲学很像吧。都是闭门造车。除非你做马克思那样的哲学哈哈哈。
@The Grand Inquisitor 天,最后一节课的内容是不是完全不会考
@Cindy Zeng 我刚在搞3040,还没开始看,最后一节你先别听吧,我当时听完倒数第二节感觉他已经把那三个问题回答的差不多了,我听完再告诉你有没有啥重点,不过听听课确实能找回一些rhythm
@The Grand Inquisitor 天,HW 13 P1 你作业里写的方法是绝对严谨的吧?感觉你和TA意思一样,就是在取的时候多用了一个消除中间的巧思……考试写你的方法应该木有问题吧?
@Cindy Zeng 嗯,他中间用那个没有必要,直接用 来算积分就行,只是这里的Riemann sum 搞出来加减相消后是和无关,稍微显得有点不自然。今天不知道咋回事,用notion又开始卡了😩
@The Grand Inquisitor 我也卡。而且/eq自动给我搞成block equation了不知道怎么调回inline
@Cindy Zeng 用shift+command+E可以直接快捷键打开inline equation,但整体打字延迟还是很高,尤其是中文
@The Grand Inquisitor 可能真的是文件太大了,有嗲担心他随时崩掉。另外我看了HW13最后一题,似乎倒数第二页写完也就只需要下个结论了?也许那张失效的图不是作业答案?
趁你睡觉的时候押题一波:
小题:
- HW8P6 记得quiz 也出过。
大题(期中前*2):
- Qz3P1 Construct a bad sequence (of indices) from a NOT-Cauchy sequence.
大题(期中后*9):
- HW13P4 积分域交集零测则积分可加和。
- HW13P5 利用continuity反证的经典,sharpened IVT.
- HW12P2 Continuity证明,定义在闭区间上的连续函数零测的证明(这在原本的简单对角线的情况下不需要但是出一道变式,把平面变成空间,把f=1变成some continuous function就可以融合HW12P5,能考到很多东西。我预感他会有变式,后来他用的动词都不是“is”而是“related to” HW problems了。那道新题应该也是某种变化比较大的变式。
- HW11P5 The first set of sufficient conditions that makes an open mapping: 1) is smooth, 2) Jacobian is non-singular . “Extremely easy exercise, but a signicant result.”
- HW8P5 NUS problem. “Try this yourself to see how much time you will take to come up with a correct and pretty proof!” It is pretty indeed.
次有可能考的:
- HW12P1 挺综合的,把Riemann Integrability 原始定义的那些概念都考到了

@The Grand Inquisitor 天,Finally那段他到底在讲什么,我看了半天都没看懂。。this result 指代的是哪个?不需要 non singular 吗。
@Cindy Zeng 我上午在高铁上,这边打起来依然很卡,我用overleaf给你押题好了






